617. Merge Two Binary Trees

问题
You are given two binary trees root1 and root2.

Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.

Return the merged tree.

Note: The merging process must start from the root nodes of both trees.

递归。将root1和root2合并到root1。
如果一个节点为null，则返回另一个节点。
否则root1的值为root1 + root2的值。
root1.left递归root1和root2的left。
root2.right递归root1和root2的right。
返回root1。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if( root1 == null ){
            return root2;
        }
        if( root2 == null ){
            return root1;
        }
        
        root1.val = root1.val + root2.val;
        root1.left = mergeTrees(root1.left,root2.left);
        root1.right = mergeTrees(root1.right,root2.right);
        
        return root1;
    }
}